## anonymous 4 years ago statics problem (:

1. anonymous

2. anonymous

im confused as to what forces are acting on points a and c.

3. anonymous

Take components

4. anonymous

okay, so i did the sum of F in the y direction.. = Ay + Cy - W - T = 0 and then the sum of F in the x direction... = Ax + Cx - Fb = 0 then i did the moment about A and got Fb = 8.9, but i am stuck there.

5. anonymous

oh and Fb is acting in the negative x direction on point B.

6. anonymous

At point A, it's the Normal reaction from Ground and the Tension from the string... at C it's just the string, is Pulley Mas-less and friction-less?

7. anonymous

i assumeee the pulley has no mass nor friction.

8. anonymous

wait so how many unknowns do i have ?

9. anonymous

This is getting confusing for me :(

10. anonymous

me too! haha. and yay '94 kids !

11. anonymous

i am not sure what direction the force on point a is acting. perpendicular to the rod or the cable?

12. anonymous

Does the answer 12.78 lb sound reasonable? I can't remember how do you do this, so I just did a bunch of stuff that seemed to be reasonable and I got that =\

13. anonymous

it does sound reasonable! but the back of the book has the answers, and evidently T = 9.08 lb ):

14. anonymous

Hmmm.. Well at first i did what you did, sum of the forces in the y direction: Ay + Cy - T - 15 = 0

15. anonymous

A is experiencing a force T, that's pulling on it at 10 degrees, so the y component of that force T that is acting on a would be sin(10) that would be equal to 0.1736T. So now 0.1736T + Cy - 1T - 15 = 0 -0.8264T + Cy - 15 = 0

16. anonymous

Now if we can substitute something like that for Cy as well we could move on, but idk what to substitute for it cause there are no degrees for point C

17. anonymous

maybe we are overlooking something. ummmmm.

18. anonymous
19. anonymous

20. anonymous

$T \times \cos 10 \times 5 \times \sin 40 = \frac{5}{2} \times \cos 40 \times 15$ T is the tension in the cable. I used concept of Torques, one Torque is in effect due to its weight (rod's weight) the other is from the horizontal component of Cable's Tension (T * cos 10). Note. I calculated Toque about A.

21. anonymous

Sum of x direction Ax - Fb = 0 Fb you said = 8.9 Ax - 8.9 = 0 Ax = 8.9 cos(10)T = 8.9 0.98T = 8.9 T = 9.08

22. anonymous

i havent learned torques, but s is something it looks like i can understand. thank you both !!

23. anonymous

That kinda looks reasonable when you look at it. I think theres no Cx because its just like a continuation of Ax or something.. idk.. i can't remember.. I took it like a year and a half ago so i can't remember.. i'm gonna be a crappy engineer lol.. can't remember much lol

24. anonymous

But at least it give the right answer.. i hope the procedure is right as well =)

25. anonymous

sike i still dont know what engineering even is and i am almost done with my first year of college. gracias yall !

26. anonymous

Don't worry, you'll figure it out =D De nada =D